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Yogita Ingle 6 years, 11 months ago
Let the terms be a – 3d, a – d, a + d, a+3d
Sum of the terms = (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20
4a = 20
a = 5
Sum of the squares of the term = (a – 3d)2 + (a – d)2 + (a + d)2 + (a + 3d)2 = 20
a2 – 6ad + 9d2 + a2 – 2ad + d2 + a2 + 2ad + d2 + a2 + 6ad + 9d2 = 120
4a2 + 20d2 = 120 – – – – – – – – – – – – (a)
Substituting a = 5 into (a)
4(52) + 20d2 = 120
100 + 20d2 = 120
d = ± 1
d = ± 1
Thus, the four numbers are:
Taking d = 1
(a – 3d), (a – d), (a + d), (a + 3d) = (5 – 3), (5 – 1), (5 + 1), (5 + 3)
= 2, 4, 6, 8
Taking d = -1
(a – 3d), (a – d), (a + d), (a + 3d)
= (5 + 3), (5 + 1), (5 - 1), (5 - 3)
= 8, 6, 4, 2
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